Answer
See below:
Work Step by Step
Convert the function in the form of $f\left( x \right)\ge 0$ or $f\left( x \right)\le 0$.
$\begin{align}
& 5x\le 2-3{{x}^{2}} \\
& 5x-2+3{{x}^{2}}\le 2-3{{x}^{2}}-2+3{{x}^{2}} \\
& 3{{x}^{2}}+5x-2\le 0
\end{align}$
The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $3{{x}^{2}}+5x-2=0$.
Now, use $\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Here,
$a=3\text{,b}=5,\text{c}=-2$
So,
$\begin{align}
& x=\frac{-\left( 5 \right)\pm \sqrt{{{5}^{2}}-4\times 3\times \left( -2 \right)}}{2\times 3} \\
& x=\frac{-5\pm \sqrt{25+24}}{6} \\
& x=\frac{-5\pm 7}{6} \\
& x=-2,\frac{1}{3}
\end{align}$
Hence, $x=-2 \text{ and }\frac{1}{3}.$
These values of x are the boundary points so locate these points on the number line as,
Now as can be seen from the above number line, $x=\frac{1}{3}$ and $x=-2$ divides the number line in three intervals as,
$\left( -\infty ,-2 \right],\left[ -2,\frac{1}{3} \right],\left[ \frac{1}{3},\infty \right)$
If $f\left( x \right)\ge 0$ or $f\left( x \right)\le 0$ then x includes the boundary point in its interval, so use big brackets for including points.
Now, test the condition for $\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0$ in interval $\left( -\infty ,-2 \right]$.
Take test point $-8$ ,
$\begin{matrix}
\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0 \\
\left( -8-\frac{1}{3} \right)\left( -8+2 \right)\overset{?}{\mathop{\le }}\,0 \\
50\overset{?}{\mathop{\le }}\,0 \\
\end{matrix}$
Thus, the condition is unsatisfied.
Now, test the condition for $\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0$ in interval $\left[ -2,\frac{1}{3} \right]$.
Take test point $0$ as,
$\begin{matrix}
\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0 \\
\left( 0-\frac{1}{3} \right)\left( 0+2 \right)\overset{?}{\mathop{\le }}\,0 \\
-\frac{2}{3}\overset{?}{\mathop{\le }}\,0 \\
\end{matrix}$
Thus, the condition is satisfied.
Now, test the condition for $\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0$ in interval $\left[ \frac{1}{3},\infty \right)$.
Take test point $5$ ,
$\begin{align}
& \left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0 \\
& \left( 7-\frac{1}{3} \right)\left( 7+2 \right)\overset{?}{\mathop{\le }}\,0 \\
& 60\overset{?}{\mathop{\le }}\,0
\end{align}$
Thus, the condition is unsatisfied.
Therefore, the interval satisfying the given inequality is $\left[ -2,\frac{1}{3} \right]$.
Thus, the solution set in the interval notation will be $\left[ -2,\frac{1}{3} \right]$.
